In Exercises 9–12, construct the indicated confidence interval...

Question

In Exercises 9–12, construct the indicated confidence intervals for (a) the population variance and (b) the population standard deviation . Assume the sample is from a normally distributed population.
c = 0.90, s^2 = 35, n = 18

c = 0.90, s^2 = 35, n = 18

Accepted Answer

Hello everyone. Let's take a look at this question together. A technician is testing the consistency of a machine that cuts sheets of cardboard. The sample variance of the thickness of 20 cardboard sheets is found to be 25 millimeters squared. Assume the thickness values are normally distributed. Construct a 90% confidence interval for both the population variance and the population standard deviation. So in order to solve this question, we have to recall how to construct a confidence interval so that we can construct a 90% confidence interval for both the population variance and the population standard deviation of the thickness of the. Sheets of cardboard where we know that the sample variance of the thickness of 20 cardboard sheets is found to be 25 millimeters squared, and we assume the thickness values are normally distributed. And so the first step in constructing our confidence interval is to look at the information that we are given, which is a confidence level of 0.90, a sample size. Of 20 and a sample variance of 25, and then we can use a formula which for a normally distributed population, the confidence interval for the population variance is given by the following formula where we know that in this formula chi square sub R and Chisquare subL are the right and left critical values from the chi square distribution. And we also know our degrees of freedom is calculated as N minus 1, so our degrees of freedom is equal to 19. And then we can determine our critical chi square values, which at a confidence level of 0.90, the left and right critical values for a degree of freedom equaling 19, which are calculated from the chi square distribution table for our degrees of freedom, which is 19, our right. The value is equal to 30.144, and our left critical value is equal to 10.117. And then we can substitute these values into the formula, which for the lower limit for the population variances, we have 20 minus 1 in parenthesis multiplied by 25 divided by 30.144, which is approximately 15.76. And then for the upper limit for the population variance, we have 20 minus 1 in parentheses multiplied by 25 divided by 10.117, and this gives us a value of approximately 46.95. And so from our calculations we can conclude that the 90% confidence interval for population variance is 16 to 47, and then the 90% confidence interval for population standard deviation is given as the square root of both of the endpoints, which is approximately 4 to 7. So we have our 90% confidence interval for the population variance and the population standard deviation. And looking at our answer choices, we can identify that the correct answer choice is answer choice A. So answer choice A is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

In Exercises 9–12, construct the indicated confidence intervals for (a) the population variance and (b) the population standard deviation . Assume the sample is from a normally distributed population.
c = 0.90, s^2 = 35, n = 18

Key Concepts

Confidence Interval

Chi-Squared Distribution

Sample Variance and Standard Deviation

Watch next

In Exercises 9–12, construct the indicated confidence intervals for (a) the population variance and (b) the population standard deviation . Assume the sample is from a normally distributed population.c = 0.90, s^2 = 35, n = 18

Key Concepts

Confidence Interval

Chi-Squared Distribution

Sample Variance and Standard Deviation

Watch next

In Exercises 9–12, construct the indicated confidence intervals for (a) the population variance and (b) the population standard deviation . Assume the sample is from a normally distributed population.
c = 0.90, s^2 = 35, n = 18